Chemistry 362 Mini-EXAM 1
Chapter 3
Monday February 18, 2008
Professor Kim R. Dunbar
NAME:
ID:
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You have 30 minutes to work this examination
Total Points on Exam is 83 points
1.
(10 pts)
a.
Draw Molecular Orbitals for O22- and F2 molecules and put the electrons in the
resulting orbitals.
σ4 *
σ4 *
π2*
π2*
2p
2p
2p
π1
π1
σ3
σ3
σ2 *
σ2 *
2s
2s
σ1
O
O
O22b.
2s
σ1
F
F
F2
What are the Bond Orders in each case?
In both cases the bond order is one.
c.
Which of the two molecules would you expect to have the longer bond? Why?
The O22- molecule has a longer bond than F2. Although these two species are
isoelectronic, the oxygen molecule has a smaller effective nuclear charge than
fluorine, so the additional two electrons (compared to O2) are not as strongly
attracted to the nuclei resulting in their probability function being more
diffuse, lessening the strength of the bond and making it longer.
2.
(10 pts)
Explain how effective nuclear charge determines the relative energies of the valence
orbitals in an atom. Use the examples of Li versus F atoms to show the relative
differences in the 2s and 2p levels in these atoms.
The trend of relative energies between s and p levels is an increase in the
energy difference between them from left to right across the table. This difference is
a direct result of the trend of increasing nuclear charge across the table. As the
effective nuclear charge (Z*) increases, the energy of the s orbital decreases due to
the increased electrostatic force of the nucleus on the s electrons. Additional
electrons filling p orbitals are shielded from the nucleus by s electron density (p
orbitals are less penetrating), so the rate of descent of the p orbital energies across
the table is not as dramatic as the s orbitals. Li and F illustrate the effect of the
relative differences in energies between s and p levels nicely. Li, which lies on the
far left of the periodic table has s and p levels which are closer together in energy
than the s and p levels in F. The effect of this difference is observed in diatomic
molecules of both elements. Li2 has mixing of the adjacent s and p levels when
determining molecular orbitals, while the s and p levels in F2 are energetically
different enough to result in no mixing.
3.
(25 pts)
Draw Lewis Dot Structures for the following molecules (be sure to include all lone pairs
of electrons). Now apply the VSEPR methods to predict the shapes of the molecules.
Tell in each case if the molecule is saturated, unsaturated or electron deficient.
a.
NHF2
b.
As(CH3)4+
c.
GeCl4
d.
PCl5
e.
OICl5
4.
(10 pts)
In the case of AB3 molecules, we used a specific approach in class to account for the σ
and the π-bonding. In very brief terms, describe the approach that was used for the BF3
molecule. THERE IS NO NEED TO DRAW ANY M.O.’s.
The atoms in this trigonal planar system are sp2 hybridized. Each major sp2
lobe on every atom contains 2 electrons, for a total of 18 electrons. 6 of these
electrons comprise the σ bonding system of the molecule, which consist of 3 bonds
one between each B and each F. The remaining 6 electrons (24 e- - 18 e-) are in the
unhybridized pz orbitals on F, which can be used to make the three group orbitals.
One of these group orbitals makes bonding and antibonding overlaps with the B pz
orbital, while the other two make non-bonding overlap with the B pz orbital.
5.
(12 pts)
Sketch all the possible overlap patterns that result from combining the orbitals on atoms
A and B. In each case, identify the symmetry (σ, π, δ) and type of molecular orbital
(bonding, non-bonding, antibonding) that your drawing corresponds to.
(Label the axes x, y, z in each case)
a.
a dxz on atom A and a dxz on atom B
b.
an s orbital on atom A and a px orbital on atom B
6.
(10 pts)
Draw the Molecular Orbitals that result
from the possible in-phase and out-of-phase
combinations of the two B sp3 orbitals and the H 1 s
orbital in the B2H6 bridge shown to the left. Use
both shading and signs to show the positive and
negative signs of the lobes.
7.
(10 pts)
The two fundamentally different Bonding Theories for Molecules presented in class are
Localized and Delocalized. Briefly describe the two approaches and how they differ.
Cite an example for which the Delocalized Theory is clearly superior in describing the
reactivity of a particular molecule and explain why it works better.
Localized theory uses the Lewis scheme to describe structure and bonding
and the Delocalized theory which describes the formation of molecular orbitals for
atomic orbitals based on atomic orbital wavefunctions. Although the localized
theory is quite powerful in it’s descriptive ability, the delocalized approach more
thoroughly accounts for more bonding phenomenon and physical properties. One
molecule that illustrates the superiority of the delocalized approach versus localized
in terms of reactivity is carbon monoxide. Carbonyl complexes of metals contain
CO bonded through the carbon end, which is contrary to what one might assume
based on electronegativity and a traditional Lewis model. The molecular orbital
approach however illustrates the energy and reactivity of the lone pair of electrons
on carbon and correctly predicts the binding of CO through carbon.